Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

const int N = 3010;
double dp[N];

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);

  dp[0] = 1;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    double p;
    cin >> p;
    for (int j = n; j >= 0; --j) {
      dp[j + 1] += dp[j] * p;
      dp[j] *= (1 - p);
    }
  }
  cout << fixed << setprecision(20) << accumulate(dp + n / 2 + 1, dp + n + 1, 0.0);

  return 0;
}